This is a question from Spivak's Calculus. Question statement:
If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a} g(x)$ exists?
I have not been able to find a counterexample, so it seems that it exists. Here is my attempt at proof, and I'm mostly interested in validity of my proof.
Following is given:
$\forall \epsilon, \exists \delta_1: 0 < |x-a| < \delta_1 \Rightarrow |f(x) – L_1| < \epsilon$
$\forall \epsilon, \exists \delta_2: 0 < |x-a| < \delta_2 \Rightarrow |f(x) + g(x) – L_2| < \epsilon$
Let $\delta_2 \le \delta_1$. If greater $\delta_2$ works, then smaller one will of course work, and if smaller one works than that inequality follows from that.
Let $L_2 – L_1 = L_3$, that is $L_2 = L_3 + L_1$. Then we substitute:
$|f(x) + g(x) – L_1 – L_3| < \epsilon$
$|(g(x) – L_3) – (L_1 – f(x))| < \epsilon$
And by the inequality $|a-b| \ge |a| – |b|$:
$|g(x) – L_3| – |L_1 – f(x)| \le |(g(x) – L_3) – (L_1 – f(x))| < \epsilon$
$|g(x) – L_3| – |f(x) – L_1| < \epsilon$
$|g(x) – L_3| < \epsilon + |f(x) – L_1|$
Since $\delta_2 \le \delta_1$, we can always choose epsilon greater than $|f(x) – L_1|$ so we can make substitution:
$|g(x) – L_3| < 2\epsilon$
And we can make $2\epsilon$ as small as we wish. This completes the proof.
Thank you for any help!
Best Answer
So it can be marked off as answered...
Do you already know that if $\lim\limits_{x\to c}F(x)$ exists and $\lim\limits_{x\to c} G(x)$ exists, then $\lim\limits_{x\to c}\bigl(F(x)-G(x)\bigr)$ exists? If so, set $F(x) = f(x)+g(x)$ and $G(x) = f(x)$ to get the desired result.
The proof is (essentially) valid; it can be streamlined a bit if you simply start from $|g(x)-L_3|$ and use the triangle inequality: \begin{align*}|g(x)-L_3| &= |g(x)-L_3 + f(x)-f(x)+L_2-L_2|\\ &= \left|\Bigl(g(x)+f(x)-L_2\Bigr)-\Bigl(f(x)-(L_2-L_3)\Bigr)\right|\\ &\leq \left|\Bigl(g(x)+f(x)\bigr)-L_2\right|+\Bigl|f(x)-L_1\Bigr|\\ &\lt 2\epsilon. \end{align*}